- #1
Logarythmic
- 281
- 0
Consider the one dimensional motion of a particle in the potential
[tex]V(x)=D(e^{-2ax}-2e^{-ax})[/tex].
I'm supposed to obtain the expressions for x as a function of time separately for the cases that the total energy E is positive, zero or negative.
I have used
[tex]\frac{1}{2}m\dot{x}^2 + V(x) - E = 0[/tex]
and got the integral
[tex]\int \sqrt{\frac{m}{2(E-V(x))}}dx[/tex]
to solve.
First, is this a correct method?
Second, how do I solve this integral?
[tex]V(x)=D(e^{-2ax}-2e^{-ax})[/tex].
I'm supposed to obtain the expressions for x as a function of time separately for the cases that the total energy E is positive, zero or negative.
I have used
[tex]\frac{1}{2}m\dot{x}^2 + V(x) - E = 0[/tex]
and got the integral
[tex]\int \sqrt{\frac{m}{2(E-V(x))}}dx[/tex]
to solve.
First, is this a correct method?
Second, how do I solve this integral?